$ \lambda_{n+m}( A \times B ) = \lambda_n(A) \cdot \lambda_m(B) $

Question

Actually I have to show two things:

$(a)$ Show that: $\mathbb{B}(\mathbb{R^n}) \bigotimes\mathbb{B}(\mathbb{R^m}) =\mathbb{B}(\mathbb{R^{n+m}}) $ $\mathbb{B}$ means Borel $\sigma$-algebra.

$(b)$ Show that $ \lambda_{n+m}( A \times B ) = \lambda_n(A) \cdot \lambda_m(B) $ for $A \in \mathbb{B}(\mathbb{R^n}) $ and $ B \in \mathbb{B}(\mathbb{R^m}) $.

So $(a)$ was no big deal. But I'm stucked in $ (b) $. Maybe we could use $(a)$ somehow?

Thank you for your help.

Answer 1

Denote $\pi$ the set of all bounded rectangles in ${\bf{R}}^{n}$. Now fix an $I\in\pi$, and consider \begin{align*} \mathcal{D}_{I}=\{B\in\mathcal{B}({\bf{R}}^{m}): |I\times B|=|I|\times|B|\}. \end{align*} Then $\pi\subseteq\mathcal{D}_{I}$. Note that the set $\pi$ is a $\pi$-system. Now we are to show that $\mathcal{D}_{I}$ is a $\lambda$-system. Clearly ${\bf{R}}^{m}\in\mathcal{D}_{I}$, and whenever $(B_{n})$ is a disjoint sequence of $\mathcal{D}_{I}$, then the union of $(B_{n})$ is still in $\mathcal{D}_{I}$.

Now, given $B\in\mathcal{D}_{I}$, assume first that $B$ is contained in a bounded rectangle $Q$, then \begin{align*} \left(I\times(Q-B)\right)\cup\left(I\times B\right)=I\times Q, \end{align*} and we have \begin{align*} |I\times(Q-B)|&=|I\times Q|-|I\times B|\\ &=|I|\times|Q|-|I|\times|B|\\ &=|I|\times(|Q|-|B|)\\ &=|I|\times|Q-B|. \end{align*} Now express ${\bf{R}}^{m}$ as the union of non-overlapping rectangles $Q_{n}$, then \begin{align*} |I\times B^{c}|&=\left|I\times\left(\bigcup_{n}(Q_{n}-B)\right)\right|\\ &=\left|\bigcup_{n}(I\times(Q_{n}-B))\right|\\ &=\sum_{n}|I\times(Q_{n}-B)|\\ &=\sum_{n}|I|\times|Q_{n}-B|\\ &=|I|\times\left|\bigcup_{n}(Q_{n}-B)\right|\\ &=|I|\times|B^{c}|, \end{align*} this shows that $B^{c}\in\mathcal{D}_{I}$, so $\mathcal{D}_{I}$ is a $\lambda$-system. By $\pi$-$\lambda$ theorem, we have $\mathcal{B}({\bf{R}}^{n})\subseteq\mathcal{D}_{I}$.

Now let $A\in\mathcal{B}({\bf{R}}^{n})$ and consider \begin{align*} \mathcal{D}_{A}=\{B\in\mathcal{B}({\bf{R}}^{m}): |A\times B|=|A|\times|B|\}. \end{align*} Now $\pi\subseteq\mathcal{D}_{A}$. With the same reasoning one can show $\mathcal{D}_{A}$ is a $\lambda$-system, and use the theorem again we get $\mathcal{B}({\bf{R}}^{n})\subseteq\mathcal{D}_{A}$.